Under enlargement scale factor (-2), the lmage of p (3,6) is (-9,12) find the center of enlargement ?

To find the center of enlargement, we need to determine the point that is equidistant from both the original point P(3,6) and its image P'(-9,12) under the scale factor of -2.

Let the center of enlargement be (x, y).

The distance between P(3,6) and the center of enlargement is given by the distance formula:
√[(x-3)^2 + (y-6)^2]

The distance between P'(-9,12) and the center of enlargement is also given by the distance formula:
√[(x-(-9))^2 + (y-12)^2]

Since the center of enlargement is equidistant from both points, we can set these two distances equal to each other:
√[(x-3)^2 + (y-6)^2] = √[(x+9)^2 + (y-12)^2]

Squaring both sides to eliminate the square root:
(x-3)^2 + (y-6)^2 = (x+9)^2 + (y-12)^2

Expanding and simplifying:
x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 18x + 81 + y^2 - 24y + 144

Rearranging terms:
-24x - 18y + 126 = 0

Solving for y in terms of x:
y = (-24x + 126) / 18
y = -4x + 7

Therefore, the center of enlargement is at (x, -4x + 7).

(-1|2,8)

(-1|2,8)